Permuations of identical items

In summary: This is strange to me.This is why you should have posted your working in the first place. You can swap the two ##A_1##'s etc. without changing the order. That's why I thought you got ##\frac{12!}{(2!)^6}##.
  • #1
mr.tea
102
12

Homework Statement


There is a book with 2 volumes. Each volume exists in 3 different languages. Each language has 2 identical copies(total of 12 books).
In how many ways we can arrange them on a shelf, with no restrictions and order of the volumes is irrelevant?

Homework Equations

The Attempt at a Solution


I thought that since we have 12 books, we can arrange them in ##\frac{12!}{(2!)^6}## but apparently it is not correct. The correct answer is ##\frac{12!}{6!}## but I am not sure why.

What am I missing?
Thank you.
 
Last edited:
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  • #2
mr.tea said:

Homework Statement


There is a book with 2 volumes. Each volume exists in 3 different languages. Each language has 2 identical copies(total of 12 books).
In how many ways we can arrange them on a shelf, with no restrictions and order is irrelevant?

Homework Equations

The Attempt at a Solution


I thought that since we have 12 books, we can arrange them in ##\frac{12!}{(2!)^6}## but apparently it is not correct. The correct answer is ##\frac{12!}{6!}## but I am not sure why.

What am I missing?
Thank you.

You're missing an attempt at the solution, so we can see where you go wrong.
 
  • #3
PeroK said:
You're missing an attempt at the solution, so we can see where you go wrong.

PS as the question is stated, I would say the answer is 1. If order is irrelevant, then you have the same 12 books on the shelf in all cases.
 
  • #4
PPS I agree with your answer! If you assume that order is relevant.
 
  • #5
PeroK said:
PPS I agree with your answer! If you assume that order is relevant.

I have just edited that the order of the volumes is irrelevant... does it add any information?
 
  • #6
mr.tea said:
I have just edited that the order of the volumes is irrelevant... does it add any information?

That doesn't make a lot of sense to me. If books are on a shelf, then the order defines the arrangement; otherwise, as I said, you only have the one set of books. It's different if two books are identical, then you can swap those without changing the arrangement. But, if you swap Volume 1 and Volume 2, by definition you have a different arrangement of books.

I'd be happy that you solved the question the sensible way. I wouldn't get too hung up on what precisely the question setter had in mind in this case.
 
  • #7
PeroK said:
That doesn't make a lot of sense to me. If books are on a shelf, then the order defines the arrangement; otherwise, as I said, you only have the one set of books. It's different if two books are identical, then you can swap those without changing the arrangement. But, if you swap Volume 1 and Volume 2, by definition you have a different arrangement of books.

I'd be happy that you solved the question the sensible way. I wouldn't get too hung up on what precisely the question setter had in mind in this case.

Well, we have 6 sets of books, each set consists of 2 identical books. Like: ##A_1,A_1;B_1,B_1;C_1,C_1;A_2,A_2;B_2,B_2;C_2,C_2;## where the 1's and 2's refer to the volume(vol1 or vol2). That's the case the question talks about. So I can't understand why my answer is not correct... Maybe it's not me?
 
  • #8
mr.tea said:
Well, we have 6 sets of books, each set consists of 2 identical books. Like: ##A_1,A_1;B_1,B_1;C_1,C_1;A_2,A_2;B_2,B_2;C_2,C_2;## where the 1's and 2's refer to the volume(vol1 or vol2). That's the case the question talks about. So I can't understand why my answer is not correct... Maybe it's not me?

This is why you should have posted your working in the first place. You can swap the two ##A_1##'s etc. without changing the order. That's why I thought you got ##\frac{12!}{(2!)^6}##.

I can't make sense of the order of volumes not being relevant. If you interpret that as ##A_1, A_1, A_2, A_2## can be in any order, then you'd just get ##\frac{12!}{(4!)^3}##.
 

Related to Permuations of identical items

What are permutations of identical items?

Permutations of identical items refer to the different ways in which a set of items can be arranged or ordered, while keeping the items themselves identical. This is different from permutations of distinct items, where the items themselves are also different.

How do you calculate the number of permutations of identical items?

To calculate the number of permutations of identical items, you can use the formula n!/(x1! * x2! * x3! * ... * xn!), where n is the total number of items and x1, x2, x3, etc. represent the number of identical items for each type. This formula takes into account the fact that identical items cannot be distinguished in the final arrangement.

Is the order of identical items important in permutations?

Yes, the order of identical items is important in permutations. Even though the items themselves are identical, the different ways in which they can be ordered can result in different permutations. This is why the formula for calculating permutations of identical items includes the factorials of each type of identical item.

Can identical items be repeated in permutations?

Yes, identical items can be repeated in permutations. In fact, permutations of identical items often involve the repetition of identical items in order to create different arrangements. This is why the formula for calculating permutations of identical items includes the factorials of each type of identical item, as repeated items will result in a different permutation.

What are some real-life examples of permutations of identical items?

Some real-life examples of permutations of identical items include arranging a set of books on a shelf, arranging a deck of cards, and arranging a set of coins in a jar. In all of these cases, the items themselves may be identical, but the different ways in which they can be ordered can result in different arrangements or permutations.

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